Question

What is the change in the pH that results from the addition of 20 ml of 0.1 M NaOH to a buffer made by combining 200 ml of Benzoic Acid $\sf( K_a = 6.4×10^{-5})$ with 100ml of 0.070 M Sodium benzoate?

( Given , log(0.75) = -0.1245 & log(2) = 0.300) ​

Answer 1

Answer:

hope it will help you

by combining 200 ml of Benzoic Acid $\sf( K_a = 6.4×10^{-5})$ with 100ml of 0.070 M Sodium benzoate.

Answer 2

Answer:

by combining 200 ml of Benzoic Acid \sf( K_a = 6.4×10^{-5})(K

a

=6.4×10

−5

) with 100ml of 0.070 M Sodium benzoate.

Explanation:

hope it helps tysm for such thanks tysm ,❤️